Research article Special Issues

On the Caputo-Hadamard fractional IVP with variable order using the upper-lower solutions technique

  • Received: 16 October 2022 Revised: 03 December 2022 Accepted: 07 December 2022 Published: 19 December 2022
  • MSC : 34A12, 34A40, 47A10

  • This paper studies the existence of solutions for Caputo-Hadamard fractional nonlinear differential equations of variable order (CHFDEVO). We obtain some needed conditions for this purpose by providing an auxiliary constant order system of the given CHFDEVO. In other words, with the help of piece-wise constant order functions on some continuous subintervals of a partition, we convert the main variable order initial value problem (IVP) to a constant order IVP of the Caputo-Hadamard differential equations. By calculating and obtaining equivalent solutions in the form of a Hadamard integral equation, our results are established with the help of the upper-lower-solutions method. Finally, a numerical example is presented to express the validity of our results.

    Citation: Zoubida Bouazza, Sabit Souhila, Sina Etemad, Mohammed Said Souid, Ali Akgül, Shahram Rezapour, Manuel De la Sen. On the Caputo-Hadamard fractional IVP with variable order using the upper-lower solutions technique[J]. AIMS Mathematics, 2023, 8(3): 5484-5501. doi: 10.3934/math.2023276

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  • This paper studies the existence of solutions for Caputo-Hadamard fractional nonlinear differential equations of variable order (CHFDEVO). We obtain some needed conditions for this purpose by providing an auxiliary constant order system of the given CHFDEVO. In other words, with the help of piece-wise constant order functions on some continuous subintervals of a partition, we convert the main variable order initial value problem (IVP) to a constant order IVP of the Caputo-Hadamard differential equations. By calculating and obtaining equivalent solutions in the form of a Hadamard integral equation, our results are established with the help of the upper-lower-solutions method. Finally, a numerical example is presented to express the validity of our results.



    Let G be a simple connected graph with vertex set V(G)={v1,v2,,vn} and edge set E(G). The eigenvalues of adjacency matrix A(G) are called the eigenvalues of G. The energy E(G) of G is defined as the sum of the absolute values of its eigenvalues of A(G), which is studied in chemistry and used to approximate the total-electron energy of a molecule [3]. The singular values of an n×m matrix M are the square roots of the eigenvalues of MM if nm or MM if n<m, where M is the transpose conjugate of M. Nikiforov [4] extended the concept of energy to all matrices and defined the energy of a matrix M, denoted by E(M), as the sum of the singular values of M. Clearly, E(A(G))=E(G).

    Estrada et al. [12] introduced the atom-bond connectivity index as

    ABC(G)=vivjE(G)di+dj2didj.

    Moreover, they introduced the atom-bond connectivity matrix (or ABC matrix for short) ABCG of G, which is correlated with the ABC index of G. The (i,j)-entry of the matrix ABCG is equal to di+dj2didj if vivjE(G) and 0 otherwise. The eigenvalues of the ABC matrix of G, denoted by μ1,μ2,,μn, are said to be the ABC eigenvalues of G. The atom-bond connectivity energy (ABC energy) of a connected graph G is defined in [8] as

    EABC(G)=ni=1|μi(G)|.

    Recently, several theoretical and computational properties of the ABC energy of graphs have been obtained, see e.g., [1,8,13]. Estrada [8] and Chen [13] gave an upper bound and a lower bound for the ABC energy in terms of the general Randić index, respectively. Ghorbani et al. [1] established some new bounds for the ABC energy. Gao and Shao [7] determined the unique tree with the minimum ABC energy. In this paper, we determine the trees with the minimum ABC energy among all trees on n vertices except the star Sn.

    A matching in a graph is a set of edges without common vertices. A k-matching is a matching consisting of k edges. Let T be a tree, M be a matching of T and Mk(T) be the set of all k-matchings of T. We define mM(T) and m(T,k) by

    mM(T)=vivjM(ABCT)2ij

    and

    m(T,k)=MMk(T)mM(T),

    respectively. By Sachs Theorem [14], the characteristic polynomial ϕABC(T,x) of the ABC matrix of a tree T can be expressed as

    ϕABC(T,x)=n2k=0(1)km(T,k)xn2k.

    Then by Coulson integral formula, we get

    EABC(T)=2π+01x2ln[1+n2k=1m(T,k)x2k]dx. (2.1)

    Let T1 and T2 be two trees on n vertices. If m(T1,k)m(T2,k) for all k, then by (2.1) we get EABC(T1)EABC(T2). Moreover, if there exists some k such that m(T1,k)>m(T2,k), then EABC(T1)>EABC(T2).

    Let T be a tree on n vertices, B=(bij) be an n×n nonnegative real symmetric matrix and ABCTB. Let M be a matching of T, mM(B)=vivjMb2ij and m(B,k)=MMk(T)mM(B). Then

    E(B)=2π+01x2ln[1+n2k=1m(B,k)x2k]dx.

    Clearly, m(T,k)m(B,k). Thus EABC(T)E(B). Moreover, if (ABCT)ij>bij for some vivjE(T), then EABC(T)>E(B). Thus we can get the following lemma.

    Lemma 2.1. Let T be a tree on n vertices and B be an n×n nonnegative real symmetric matrix. If ABCTB, then EABC(T)E(B). Moreover, if (ABCT)ij>bij for some vivjE(T), then EABC(T)>E(B).

    Let uv be an edge of a tree T and Tuv=T1T2, where T1(T2) is the component of Tuv containing u(v, respectively). We denote the sub-matrices of ABCT spanned by the vertices of T1 and T2 by (ABCT)V(T1) and (ABCT)V(T2), respectively.

    By Lemma 2.1, we have the next lemma.

    Lemma 2.2. Let uv be an edge of a tree T and Tuv=T1T2, where T1(T2) is the component of Tuv containing u(v, respectively). Then

    EABC(T)>E((ABCT)V(T1))+E((ABCT)V(T2)).

    Suppose that uv is not a pendent edge. If dT(w)2 for any wNT(u){v}, then

    E((ABCT)V(T1))EABC(T1).

    Furthermore, if d(w)2 for any wNT(u)NT(v){u,v}, then

    EABC(T)>EABC(T1)+EABC(T2).

    Lemma 2.3. ([7]) Let T be a tree of order n3. Then EABC(T)2n2, withequality if and only if TSn, where Sn is the star of order n.

    Lemma 2.4. ([7]) Let t2,xi3 for i=1,,t, and ti=1xi8. Then ti=2xi2ti=2xi+(t1)2.

    For two graphs G and H, we define GH to be their disjoint union. In addition, let kG be the disjoint union of k copies of G. Let Sn be the tree formed by attaching a vertex to a pendent vertex of the star Sn1. Note that

    ϕABC(Sn,x)=xn4[x4(1+(n3)2n2)x2+(n3)22(n2)].

    Thus

    EABC(Sn)=2n3+1n2+2n4+1n2.

    Lemma 3.1. Let x11. Then

    x5+1>x3+1x2+2x4+1x2. (3.1)

    Proof. It is equivalent to prove that

    2x52x4+1x21x21>0. (3.2)

    Let f(x)=2x52x4+1x21x21 with x11. Then

    dfdx=1x5221x4+1x2(11(x2)2)+1(x2)2>1x5221x4+1x2=1x512(x4)+2x2>0.

    Thus f(x) is a strictly monotonously increasing function on x. Noting that f(11)=0.0166>0, then the lemma holds.

    From Lemma 3.1, EABC(Sn)<2+2n5 for n11.

    For n=1,2,3, there is only unique tree Sn. For n=4, there are exactly two trees P4 and S4. Obviously, P4 is the tree with the second minimum ABC energy. For n=5, there are exactly three trees P5, S5 and S5. By direct calculation, we have EABC(S5)=3.9831>EABC(P5)=2+6>23=EABC(S5). Thus P5 is the tree with the second minimum ABC energy. Let P5=v1v2v3v4v5, we denote the tree, obtained by attaching a new vertex to v2 of P5, by P6. For n=6, there are exactly six trees T2.8,T2.9,T2.10,T2.11,T2.12,T2.13 (see tables of graph spectra in [14]), where T2.8S6,T2.9S6, T2.11P6 and T2.13P6. By direct calculation, EABC(T2.12)=5.0590>EABC(P6)=4.9412>EABC(T2.10)=4.8074>EABC(S6)=4.6352>EABC(P6)=4.6260>EABC(S6)=4.

    By simple calculations, we obtain the following lemma.

    Lemma 3.2. Let T be an n-vertex tree not isomorphic to Sn, where 7n10. Then EABC(T)EABC(Sn) with equality if and only if TSn.

    Lemma 3.3. Let T be a tree on n11 vertices.

    (i) Let u1v1E(G) and Tu1v1=T1T2, where T1(T2) is the component of Tu1v1 containing u1(v1, respectively). If d(w)2 for any wN(u1)N(v1){u1,v1} and |V(T1)|=n1|V(T2)|=n23, then EABC(T)>EABC(Sn).

    (ii) Let u2v2,u3v3E(G), Tu2v2P2T3 and T3u3v3P2T4, where T3 is one of the component of Tu2v2 and T4 is one of the component of T3u3v3. If dT(w1)2 for any w1NT(u2)NT(v2){u2,v2} and dT3(w2)2 for any w2NT3(u3)NT3(v3){u3,v3}, then EABC(T)>EABC(Sn).

    Proof. (ⅰ) By Lemmas 2.2 and 2.3, we have

    EABC(T)>EABC(T1)+EABC(T2)2n12+2n222n5+2>EABC(Sn).

    (ⅱ) Similarly, by Lemmas 2.2 and 2.3, we have

    EABC(T)>EABC(T3)+E((ABCT)V(P2))>EABC(T4)+2+22n6+222n5+2>EABC(Sn).

    We complete the proof.

    Lemma 3.4. Let n11. Then EABC(Pn)>EABC(Sn).

    Proof. By Lemma 2.2, we have EABC(Pn)>EABC(P3)+EABC(Pn3)2+2n5>EABC(Sn).

    A tree is called starlike if it has exactly one vertex of degree greater than two.

    Lemma 3.5. Let TSn be a starlike tree with order n11 and v be the unique vertex with degree at least three. Let Tv=n1P1n2P2nmPm and mi=1ini+1=n. Then EABC(T)>EABC(Sn).

    Proof. If ni1 for some i3, then there exists an edge uv such that Tuv=T1T2, where T1(T2) is the component of Tuv containing u(v, respectively), |V(T1)|,|V(T2)|3 and d(w)2 for any wN(u)N(v){u,v}. Then by (ⅰ) of Lemma 3.3 we can get the result. Suppose that ni=0 for all i3. If n2=0, then TSn. If n2=1, then TSn. If n22, then by (ⅱ) of Lemma 3.3 we can get the result.

    Let T be a tree and R(T) be set of vertices of degree greater than two in T.

    Lemma 3.6. Let T be a tree with n11 vertices and |R(T)|2. If there are no adjacent vertices in R(T), then EABC(T)>EABC(Sn).

    Proof. Let d(u)3 and d(v)3 and Pl=uv1vl1v be the single path connecting u and v with d(v1)==d(vl1)=2. Clearly, l2. Without loss of generality, we suppose that Tuv1=T1T2 such that T1 is a starlike-tree or a path, where uV(T1).

    If l3, then by (i) of Lemma 3.3 we can get the result.

    Suppose now that l=2. Let T1(T2) be the component of Tuv1v1v containing u(v, respectively), s=|V(T1)| and t=|V(T2)|. Obviously, s+t+1=n.

    If s=3, then the ABC matrix of T can be written as

    ABCT=[BCCD],

    where

    B=[0023000230232301200120],C=[03×103×(t1)1201×(t1)],

    and D=(ABCT)V(T2). Let

    A=[B00ABCT2].

    Obviously, DABCT2. Thus ABCT>A. By Lemmas 2.1 and 2.3, we have

    EABC(T)>E(A)=E(B)+EABC(T2)21+13+12+2n62n5+2>EABC(Sn).

    Suppose that s=4. Then T1S4 or P4. If T1S4, then the ABC matrix of T can be written as

    ABCT=[FHHK],

    where

    F=[000340000340000340343434012000120],H=[04×104×(t1)1201×(t1)],

    and K=(ABCT)V(T2). Let

    M=[F00ABCT2].

    Obviously, KABCT2. Thus ABCT>M. By Lemmas 2.1 and 2.3 we have

    EABC(T)E(M)=E(F)+EABC(T2)22+14+12+2n72n5+2>EABC(Sn).

    Suppose now that T1P4. Let T1=u1uu2u3. Then by Lemmas 2.2 and 2.3, we have

    EABC(T)2+EABC(Tu2u3)2+EABC(P3)+EABC(T2)2+2+2n72n5+2>EABC(Sn).

    By symmetry, we now suppose that 5s,tn6, then by Lemmas 2.2 and 2.3, we have

    EABC(T)>EABC(T1)+EABC(T2)2s2+2t223+2n8>2+2n5>EABC(Sn).

    Lemma 3.7. Let T be a tree with n11 vertices and |R(T)|2. If there exist adjacent vertices in R(T), then EABC(T)>EABC(Sn).

    Proof. Let E0={uvE(T)|d(u),d(v)3}, and TE0=xP1yP2T1Tz, where T1,,Tz are components of TE0 with at least three vertices. Let xP1={v1,,vx} and yP2={vx+1vx+2,,vx+2y1vx+2y}. Then dT(vi)3 with 1ix, dT(vx+2j1)3 and dT(vx+2j)=1 with 1jy, and for each component Ti with 1iz, there exists a vertex viV(Ti) such that dT(vi)dTi(vi)+1. Let |V(Ti)|=si for 1iz. Thus we have

    2(n1)=vV(T)dT(v)3x+4y+zi=1(vV(Ti)dTi(v)+1)=3x+4y+zi=12(si1)+z=x+2nz.

    Thus we get that zx+2. We discuss the following four cases.

    Case 1. y=0 and z=2.

    Then x=0 and s1+s2=n. By Lemmas 2.2 and 2.3, we get

    EABC(T)>EABC(T1)+EABC(T2)2s12+2s222n5+2>EABC(Sn).

    Case 2. y=0 and z3.

    Then x+zi=1si=n. Without loss of generation, we suppose that 3szsz1s2s1.

    If z1i=1si=6, then z=3,sz=3 and x1. Thus n10, a contradiction.

    If z1i=1si=7, then z=3,s1=4,s2=s3=3. Thus x=1 and n=11. Obviously, T2T3S3 and T1S4 or P4. By Lemmas 2.2 and 2.3, we have

    EABC(T)EABC(T1)+E((ABCT)V(T2))+E((ABCT)V(T3))242+4×23=7.448>6.8742=EABC(S11).

    Suppose that z1i=1si8. By Lemmas 2.2–2.4, we have that

    EABC(T)>zi=1EABC(Ti)2zi=1si22z1i=1si+(z1)3+2sz22nxsz+x+24+2sz2=2nsz2+2sz22n5+2>EABC(Sn). (3.3)

    Case 3. y1 and z3.

    Then z1i=1si+sz+x+2y=n. By Lemmas 2.2 and 2.3, we have

    EABC(T)2y23+zi=1EABC(Ti)2y23+2zi=1si2.

    If z1i=1si8, then by Lemma 2.4, we have

    EABC(T)2y23+2z1i=1si+(z11)2+2sz22y23+2nsz2yx+x+24+2sz2=2y23+2nsz2y2+2sz22y23+2n32y2+2322n5+2>EABC(Sn).

    Here, the last but one inequality holds because f(y)=2y23+2n52y+2 is increasing for 0y2n134.

    Suppose that z1i=1si7. Then z=3. Thus s1+s2=6 or 7, s3=3 and x1.

    Suppose first that s1+s2=7 and s3=3. If x=0, then n=2y+10. Hence

    EABC(T)2y23+EABC(T1)+EABC(T2)+EABC(T3)2y23+22+2+2=(n10)23+22+42n5+2>EABC(Sn).

    Suppose now that x=1. Then n=2y+11. Hence

    EABC(T)>2y23+22+2+2=(n11)23+22+4.

    Let f(x)=(x11)23+4+222x3+1x2+2x4+1x2. It is easy to get that f(x)>0 for x11. Then f(x) is an increasing function on x and f(x)f(11)>0. Thus

    EABC(T)>(n11)23+4+22>2n3+1n2+2n4+1n2.

    By a similar discussion as above, we can get the result for the case s1+s2=6 and s3=3.

    Case 4. y1 and z=2.

    Then x=0,n=2y+s1+s2. If n2y11, then by Lemmas 2.1–2.3, we have

    EABC(T)2y23+EABC(T1)+EABC(T2)2y23+2s12+2s222y23+2n2y32+2322n5+2>EABC(Sn).

    Suppose that n2y10. Then 6s1+s210. If s1+s2=10, then by Lemmas 2.1–2.3, we have

    EABC(T)>2y23+232+272=(n10)23+2+252n5+2>EABC(Sn).

    Similarly, for each 6s1+s29, we may also get the result.

    Combining Lemmas 3.2 and 3.4–3.7, we get our main result.

    Theorem 3.1. Among all trees (except the star) on n5 vertices, P5 is the unique tree with the minimum ABC energy for n=5, P6 is the unique tree with the minimum ABC energy for n=6 and Sn is the unique tree with the minimum ABC energy for n7.

    In this paper, motivated by the unique tree with the minimum ABC energy, we determine the trees with the minimum ABC energy among all trees on n vertices except the star Sn.

    The authors would like to thank anonymous referees for helpful comments and suggestions which improved the original version of the paper. This work was supported by the Natural Science Foundation of Guangdong Province (No.2021A1515010028).

    The authors declare that they have no competing interests.



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